Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x+8y &= -3 \\ -6x+5y &= 3\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-6x = -5y+3$ Divide both sides by $-6$ to isolate $x$ $x = {\dfrac{5}{6}y - \dfrac{1}{2}}$ Substitute this expression for $x$ in the first equation. $6({\dfrac{5}{6}y - \dfrac{1}{2}}) + 8y = -3$ $5y - 3 + 8y = -3$ Simplify by combining terms, then solve for $y$ $13y - 3 = -3$ $13y = 0$ $y = 0$ Substitute $0$ for $y$ in the top equation. $6x+8( 0) = -3$ $6x = -3$ $6x = -3$ $x = -\dfrac{1}{2}$ The solution is $\enspace x = -\dfrac{1}{2}, \enspace y = 0$.